NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

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In this chapter, the complex numbers are explained, their properties, operations on complex numbers along with quadratic equations.

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NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

Edited By Ravindra Pindel | Updated on Sep 12, 2022 05:49 PM IST

NCERT Exemplar Class 11 Maths solutions chapter 5 covers Complex numbers and its applications. Complex numbers are something which we have been studying over the years with any ‘real number’ being said to be a complex number, which can be denoted as a variable or alphabet in simpler terms. The other number, along with the real number, is a part of the complex number, is called the ‘imaginary number.’ Both of them appear together in a problem. Terms such as integer, conjugate, square root, polar, etc. will be studied in NCERT Exemplar Class 11 Maths solutions chapter 5, that will be utilised to find results for these complex number problems.

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NCERT Exemplar Class 11 Maths Solutions Chapter 5: Exercise - 1.3
Important Notes from NCERT Exemplar Class 11 Maths Solution chapter 5
Main Subtopics in NCERT Exemplar Class 11 Maths Solution Chapter 5
What will the students learn from NCERT Exemplar Class 11 Maths Solution Chapter 5?
NCERT Solutions for Class 11 Mathematics Chapters
Important Topics in NCERT Exemplar Class 11 Maths Solutions Chapter 5

NCERT Exemplar Class 11 Maths Solutions Chapter 5: Exercise - 1.3

Question:1

For a positive integer n, find the value of $(1 - i)^n$ $(1-1/i)^n$

Answer:

$(1-i)^n (1-1/i)^n$
$=(1-i)^n (1+i )^n$
$=(1-i^2 )^n$
$=2^n$

Question:2

Evaluate $\sum_{i=1}^{13}$ $(i^n+i^{n+1} )$ where n $\epsilon$ N.

Answer:

= $\sum_{i=1}^{13}$ $(i^n+i^{n+1} )$
$=$ $\sum_{i=1}^{13}$ $(1+i) i^n$
$=(1+i)(1+i^2+i^3+i^4....+i^{12}+i^{13} )$
$=(1+i) \frac{i(i^{13}-1)}{(i-1)}$
$=(1+i) i\frac{(i-1)}{(i-1)}$
$=(1+i)i$
$=i+i^2$
$=i-1$

Question:3

If $\frac{(1+i)}{(1-i)}^3-\frac{(1-i)}{(1+i)}^3= x+iy$ then find (x, y).

Answer:

Given $\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3= x+iy$
$=\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3$
$=\left (\frac{1+2i+i^2}{1-i^2} \right )^{3}-\left (\frac{1-2i+i^2}{1-i^2} \right )^{3}$
$=(\frac{2i}{2})^3-(\frac{-2i}{2})^3$
$=i^3-(-i^3 )$
$=2i^3=0-2i$
$Thus,(x,y)=(0,-2)$

Question:4

If $(1+i)^2/(2-i)= x+iy$ then find the value of x + y.

Answer:

$x+iy=\frac{(1+i)^2}{(2-i)}=\frac{ (1+2i+i^2)}{(2-i)}= \frac{2i}{(2-i)}$

$On\: \: rationalising\: \: 2i(2+i)/(2-i)(2+i)$ ,
$=\frac{(4i+2i^2)}{(4-i^2 )}=\left (\frac{(4i-2)}{(4+1)} \right )$
$=\frac{-2}{5}+\frac{4i}{5}$
$x=\frac{-2}{5} \: \: and\: \: y=\frac{4}{5}$

$So, x+y= -\frac{2}{5}+\frac{4}{5}=\frac{2}{5}$

Question:5

If $\left(\frac{1-i}{1+i}\right)^{100}=a+ib$ then find (a, b).

Answer:

$a+ib=\left(\frac{1-i}{1+i}\right)^{100}=\left [\frac{1-i}{1+i}*\frac{1-i}{1-i} \right ]^{100}$
$=\left [\frac{(1-i)^2}{1-i^2} \right ]^{100}$
$=\left (\frac{(1-2i+i^2)}{(1+1)} \right )^{100}$
$=\left (\frac {-2i}{2 }\right )^{100}$
$=(i^4 )^{25}=1$
$Hence,(a,b)=(1,0)$

Question:6

If $a=\cos \theta+i \sin\theta$ , find the value of $\frac{(1+a)}{(1-a)}$ .

Answer:

$a=\cos \theta+i \sin\theta$
$\frac{(1+a)}{(1-a)}$ $=\frac{(1+cos\theta)+i \sin \theta}{(1-cos\theta)-isin\theta}$

$=\frac{\left ( 2\cos ^{2}\frac{\theta }{2} +i2\sin\frac{\theta }{2}\cos\frac{\theta }{2} \right )}{\left ( 2\sin ^{2}\frac{\theta }{2} -i2\sin\frac{\theta }{2}\cos\frac{\theta }{2} \right )}$
$=\frac{2\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{2\sin\frac{\theta }{2}\left ( \sin\frac{\theta }{2}-i\cos\frac{\theta }{2} \right )}$
$=\frac{i\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{\sin\frac{\theta }{2}\left ( i\sin\frac{\theta }{2}-i^2\cos\frac{\theta }{2} \right )}$
$=\frac{i\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{\sin\frac{\theta }{2}\left ( i\sin\frac{\theta }{2}+\cos\frac{\theta }{2} \right )}$
$=i\cot \frac{\theta }{2}$

Question:7

If $(1 + i)z = (1 - i) \bar{z},$ then show that $z= -i\bar{z}$

Answer:

$(1 + i)z = (1 - i) \bar{z},$
$z=\frac{1-i}{1+i}\bar{z}$
Rationalising the denominator,
$\frac{\left (1-i \right )\left (1-i \right )}{\left (1+i \right )\left (1-i \right )}\bar{z}$
$=\frac{\left ( 1-i \right )^{2}}{1-i^2}\bar{z}$
$=\frac{\left ( 1-2i+i^2 \right )}{1+1}\bar{z}$
$=\frac{\left ( 1-2i-1 \right )}{2}\bar{z}$
$=-i\bar{z}$
Hence,Proved.

Question:8

If $z = x + iy$ , then show that $z\bar{z}+2(z+\bar{z})+b=0$ where b?R, representing z in the complex plane is a circle.

Answer:

$z=x+iy$
$\bar{z}=x-iy$
$Now,we\: \: also\: \: have\: \: ,z\bar{z }+2(z+\bar{z})+b=0$
$(x+iy)(x-iy)+2(x+iy+x-iy)+b=0$
This equation is a equation of circle

Question:9

If the real part of $\frac{(\bar{z}+2)}{(\bar{z}-1)}$ is 4, then show that the locus of the point representing z in the complex plane is a circle.

Answer:

$Let \: \: z=x+iy \: \:$
$\frac{\bar{z}+2}{\bar{z}-1}= \frac{x-iy+2}{x-iy-1}$
$=\frac{\left [ \left ( x+2 \right )-iy \right ]\left [ \left ( x-1 \right )+iy \right ]}{\left [ \left ( x-1 \right )-iy \right ]\left [ \left ( x-1 \right )+iy \right ]}$
$=\frac{(x-1)(x+2)+y^2+i[(x+2)y-(x-1)y]}{(x-1)^2+y^2}$
$real \: \: part=4 \Rightarrow \frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4$
$x^2+x-2+y^2=4(x^2-2x+1+y^2 )$
$3x^2+3y^2-9x+6=0$

The equation obtained is that of a circle. Hence, locus of z is a circle.

Question:10

Show that the complex number z, satisfying the condition $arg\left ( \frac{z-1}{z+1} \right )=\frac{\pi}{4}$ lies on a circle.

Answer:

Let z=x+iy
$arg\left ( \frac{z-1}{z+1} \right )=\frac{\pi}{4}$
$arg\left ( z-1 \right )-arg\left ( z+1 \right )=$ $\frac{\pi}{4}$
$arg\left ( x+iy-1 \right )-arg\left ( x+iy+1 \right )=$ $\frac{\pi}{4}$
$arg\left ( x-1+iy \right )-arg\left ( x+1+iy \right )=$ $\frac{\pi}{4}$
$\tan^{-1}\frac{y}{x-1}-\tan^{-1}\frac{y}{x+1}=$ $\frac{\pi}{4}$
$\tan^{-1}\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\left (\frac{y}{x-1} \right )\left (\frac{y}{x+1} \right )}=$ $\frac{\pi}{4}$
$\tan^{-1}\left(\frac{y\left ( x+1+-x+1 \right )}{x^2-1+y^2}\right) = \frac{\pi}{4}$
$\tan\frac{\pi}{4} = \frac{2y}{x^2+y^2-1}$
$x^2+y^2-1=2y$
$x^2+y^2-1-2y=0$
The equation obtained represents the equation of a circle

Question:11

Solve that equation $|z| = z + 1 + 2i$ .

Answer:

$|z| = z + 1 + 2i$
$Substituting \: \: z=x+iy \: \: we \: \: get \: \: |x+iy|=x+iy+1+2i$
$|z|= \sqrt{(x^2+y^2 )} =(x+1)+i(y+2)$
Comparing real and imaginary parts $\sqrt{(x^2+y^2 )} =(x+1)$
And $0=y+2 , y=-2$
Substituting the value of y in $\sqrt{(x^2+y^2 )} =(x+1)$
$x^2+(-2)^2=(x+1)^2$
$x^2+4=x^2+2x+1$

Hence, $x=\frac{3}{2}$
Hence, $z=x+iy$
$=\frac{3}{2} -2i$

Question:12

If $|z + 1| = z + 2 (1 + i)$ , then find z

Answer:

We have $|z + 1| = z + 2 (1 + i)$
Substituting $z=x+iy$ we get $x+iy+1=x+iy+2i+1$
$|z|= \sqrt{(x^2+y^2 ) }=\sqrt{(x+1)^2+y^2 } = (x+2)+i(y+2)$
Comparing real and imaginary parts, $\sqrt{(x+1)^2+y^2 } = (x+2)$
$and 0=y+2 ;y=-2$
Substituting the value of y in $\sqrt{(x+1)^2+y^2 } = (x+2)$
$(x+1)^2+(-2)^2=(x+2)^2$
$x^2+2x+1+4=x^2+4x+4$
$2x=1$
Hence, $x=\frac{1}{2}$
Thus, $z=x+iy=\frac{1}{2}-2i$

Question:13

If $arg (z - 1) = arg (z + 3i)$ , then find x – 1 : y. where $z = x + iy$

Answer:

Given that $arg (z - 1) = arg (z + 3i)$
$arg (x+iy-1)=arg(x+iy+3i)$
$arg(x-1+iy)=arg(x+i(y+3))$
$\tan^{-1}\frac{y}{x-1}=\tan^{-1}\frac{y+3}{x}$
$\frac{y}{x-1}=\frac{y+3}{x}$
$xy=xy-y+3x-3$
$3x-3=y$
$\frac{x-1}{y}=\frac{1}{3}$

Question:14

Show that $\left | \frac{z-2}{z-3} \right |=2$ represents a circle. Find its centre and radius.

Answer:

$\left | \frac{z-2}{z-3} \right |=2$ Substituting $z=x+iy$ , we get $\left | \frac{x+iy-2}{x+iy-3} \right |=2$
$|x-2+iy|=2|x-3+iy|$
$\sqrt{(x-2)^2+y^2 }=2\sqrt{((x-3)^2+y^2}$
$x^2-4x+4+y^2=4(x^2-6x+9+y^2 )$
$3x^2+3y^2-20x+32=0$
$x^2+y^2-\frac{20}{3} x+\frac{32}{}3=0$
$(x-\frac{10}{3})^2+y^2+\frac{32}{3}-\frac{100}{9}=0$
Thus, the centre of circle is $(\frac{10}{3},0)$ and radius is $2/3$

Question:15

If $\frac{z-1}{z+1}$ is a purely imaginary number $(z \neq -1)$ , then find the value of $|z|$ .

Answer:

Let $z=x+iy$
$\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}$
$\frac{\left [ \left ( x-1 \right ) +iy\right ]\left [ \left ( x+1 \right )-iy \right ]}{\left [ \left ( x+1 \right ) +iy\right ]\left [ \left ( x+1 \right )-iy \right ]}$
$=\frac{(x-1)(x+1)+y^2+i[(x+1)y-(x-1)y]}{(x+1)^2+y^2 }$
$\frac{z-1}{z-1}$ is purely imaginary
$\frac{(x-1)(x+1)+y^2}{(x+1)^2+y^2 }=0$
$x^2-1+y^2=0$
$x^2+y^2=1$ Hence, $|z|=1$

Question:16

z₁ and z₂ are two complex numbers such that $|z_1| = |z_2|$ and $arg (z_1) + arg (z_2) = \pi$ , then show that $|z_1| = -|z_2|$ .

Answer:

$\theta_1+\theta_2=\pi$

$\theta_1=\pi-\theta_2$

$z_2=\left |z_2 \right |\left (cos\theta _2+i\ sin\theta_2 \right )$

$z_1=|z_2 |\left ( -\cos \theta_2+isin\theta_2 \right )$

$z_1=-|z_2 |\left ( \cos \theta_2-isin\theta_2 \right )$

$z_1=-|z_2 |\left ( \cos \theta_2-isin\theta_2 \right )$ , $|z_1| = -|z_2|$

Question:17

If |z₁| = 1 (z₁ ≠ –1) and $z_2= \frac{z_1-1}{z_1+1}$ then show that the real part of z₂ is zero

Answer:

Let $z_1=x+iy$ $|z_1 |= \sqrt{x^2+y^2} =1$
$z_2= \frac{z_1-1}{z_1+1}=\frac{x+iy-1}{x+iy+1}$
$=\frac{[(x-1)+iy][(x+1)-iy]}{[(x+1)+iy][(x+1)-iy] }$
$= \frac{(x-1)(x+1)+y^2+i[(x+1)y-(x-1)y]}{(x+1)^2+y^2}$
$\frac{x^2+y^2-1+2iy}{(x+1)^2+y^2 }=0$
Since, $x^2+y^2=1$
$\frac{1-1+2iy}{(x+1)^2+y^2}=\frac{0+2iy}{(x+1)^2+y^2}$
Therefore, the real part of $z_2$ is zero

Question:18

If z₁, z₂ and z₃, z₄ are two pairs of conjugate complex numbers, then find $arg\left ( \frac{z_1}{z_4} \right )+arg\left ( \frac{z_2}{z_3} \right )$

Answer:

$z_1$ and $z_2$ are conjugate complex numbers.
The negative side of the real axis $= r_1 (cos\theta-isin \theta)$
$=r_1\left ( \cos\left ( -\theta_1 \right )+i\sin\left (- \theta_1 \right ) \right )$
Similarly, $z_3=r_2\left ( \cos\left ( \theta_2 \right )-i\sin\left ( \theta_2 \right ) \right )$
$z_4=r_2\left ( \cos\left ( -\theta_2 \right )+i\sin\left (- \theta_2 \right ) \right )$
$arg \left ( \frac{z_1}{z_4} \right )+arg \left ( \frac{z_2}{z_3} \right )$ $=arg(z_1)-arg(z_4)+ arg(z_2)- arg(z_3)??$
$=\theta_1-(-\theta_2)+(-\theta_1)-\theta_2=0$

Question:19

If $|z_1| = |z_2| =... = |z_n| = 1$ , then
Show that $|z_1 + z_2 + z_3 +... + z_n| =|1/z_1 +1/z_2 +...+1/z_n |$

Answer:

$|z_1| = |z_2| =... = |z_n| = 1$
$|z_1|^2 = |z_2|^2 =... = |z_n|^2 = 1$
$z_1 \bar{z_1 } =z_2 \bar{z_2 }=...=z_n \bar{z_n }$
$z_1=\frac{1}{\bar{z_1}},z_2=\frac{1}{\bar{z_2}},...z_n=\frac{1}{\bar{z_n}}$
Now, $|z_1+z_2+z_3+...z_n |$ $=\left | \frac{z_1\bar{z_1}}{\bar{z_1}}+\frac{z_2\bar{z_2}}{\bar{z_2}}+...+\frac{z_n\bar{z_n}}{\bar{z_n}} \right |$
= $\left | \frac{1}{\bar{z_1}}+\frac{1}{\bar{z_2}}+...+\frac{1}{\bar{z_n}} \right |$
$=\left | {\frac{1}{z_1}+\frac{1}{z_2}+...+\frac{1}{z_n}} \right |$

Question:20

If for complex numbers z₁ and z₂, $arg (z_1) - arg (z_2) = 0$ , then show that $\left | z_1-z_2 \right |= |z_1|-|z_2|$

Answer:

Let $z_1=\left | z_1 \right |\left ( \cos\theta_1 +i\sin \theta_1 \right )$ and $z_2=\left | z_2 \right |\left ( \cos\theta_2 +i\sin \theta_2 \right )$
$arg(z_1 )- arg(z_1 )=0$
$\theta _1-\theta _2=0$
$\theta _1=\theta _2$
$z_1-z_2= (z_1 |cos\theta_1-| z_2|cos\theta_1 )+i(z_1 |sin\theta_1-| z_2|sin\theta_1 )$
$z_1-z_2= \sqrt{(z_1 |cos\theta_1-| z_2|cos\theta_1 )^{2}+(z_1 |sin\theta_1-| z_2|sin\theta_1 )^{2}}$
$=\sqrt{|z_1|^2+|z_2|^2-2|z_1||z_2|\cos^2\theta_1-2|z_1||z_2|\sin^2\theta_1}$
$=\sqrt{|z_1|^2+|z_2|^2-2|z_1||z_2|}$
$=\sqrt{\left (|z_1|-|z_2| \right )^2}= |z_1|-|z_2|$

Question:21

Solve the system of equations $Re (z^2) = 0, |z| = 2.$

Answer:

$Re (z^2) = 0, |z| = 2.$
$Let\: \: z=x+iy$
$|z|= \sqrt{x^2+y^2 } =2$
${x^2+y^2 } =4$
$z^2=x^2+2ixy-y^2$
$=(x^2-y^2 )+2ixy$
$Now,Re (z^2 )=0$
$x^2-y^2=0$
$x^2=y^2=2$
$x=y=\pm \sqrt{2}$
Hence, $z=x+iy=\pm\sqrt{2}\pm i\sqrt{2}$
$=\sqrt{2}+i\sqrt{2},\sqrt{2}-i\sqrt{2},-\sqrt{2}+i\sqrt{2} \: \: and\: \: -\sqrt{2}-i\sqrt{2}$

Question:22

Find the complex number satisfying the equation $z+\sqrt{2} |(z+1)|+i=0$ .

Answer:

$z+\sqrt{2} |(z+1)|+i=0\: \: \: \: ...(i)$
Substituting $z=x+iy$ we get $x+iy+\sqrt2 |x+iy+1|+i=0$
$x+i(1+y)+\sqrt2 \left [ \sqrt{(x+1)^2+y^2 } \right ]=0$
$x+i(1+y)+\sqrt2 \left [ \sqrt{(x^2+2x+1+y^2 ) } \right ]=0$
1+y = 0
y = -1
$x+\sqrt2 \sqrt{x^2+2x+2}=0$
$\sqrt2 \sqrt{x^2+2x+2}=-x$
$2x^2+4x+4=x^2$
$x^2+4x+4=0$
$(x+2)^2=0$
$x=-2$
Hence, $z=x+iy=-2-i$

Question:23

Write the complex number $z=\frac{1-i}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$ in polar form

Answer:

$z=\frac{1-i}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$
$=\frac{\sqrt{2}\left [ \frac{1}{\sqrt{2}}-i1/\sqrt{2} \right ]}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$
$=\frac{\sqrt{2}\left [\cos\frac{\pi}{4}-isin\frac{\pi}{4} \right ]}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$
$=\sqrt2\left [ \cos \left ( -\frac{\pi}{4}-\frac{\pi}{3}\right )+i\sin\left (-\frac{\pi}{4}-\frac{\pi}{3} \right ) \right ]$
$=\sqrt2\left [ \cos \left ( -\frac{7\pi}{12}\right )+i\sin\left (-\frac{7\pi}{12} \right ) \right ]$
$=-\sqrt2\left [ \cos \left ( \frac{5\pi}{12}\right )+i\sin\left (\frac{5\pi}{12} \right ) \right ]$

Question:24

If z and w are two complex numbers such that $|zw| = 1$ and $arg(z) - arg (w) = \pi/2$ , then show that $zw=-i$ .

Answer:

Let $z=\left | z \right |(\cos\theta_1 + i sin\theta_1 )$ and $w=\left | w \right |(\cos\theta_2 + i sin\theta_2 )$
$|zw|=|z||w|=1$
Also $arg(z)-arg(w)= \frac{\pi}{2}$
$\theta_1-\theta_2=\frac{\pi}{2}$
$\bar{z} w=|z|\left (\cos \theta _{1}-i\sin\theta _{1} \right )w$
$w=|w|\left (\cos \theta _{2}+i\sin\theta _{2} \right )=1$
$\bar{z}w$ $=|z||w|(\cos (-\theta_1)+i sin(-\theta_1 ))(cos\theta_2+i \sin\theta_2 )$
$\bar{z}w$ $=\cos [(\theta _2 - \theta _1 )+isin( \theta _2- \theta _1 )]$
$=\left [ \cos \left ( -\frac{\pi}{2} \right ) +i\sin\left ( -\frac{\pi}{2} \right )\right ]$
$=1[0-i]=-i$

Question:25

i) For any two complex numbers z₁, z₂ and any real numbers a, b, $|az_1 - bz_2|^2 + |bz_1 + az_2|^2 =....$

ii) The value of $\sqrt{-25} * \sqrt{-9}$ is

iii) The number $\frac{(1-i)^3}{1-i^3}$ is equal to .......

iv) The sum of the series $i+i^2+i^3+...+i^{1000}$ upto 1000 terms is ...

v) Multiplicative inverse of 1 + i is ................

vi) If $z_1 \ and \ z_2$ are complex numbers such that $z_1 + z_2$ is a real number, then $z_2$ =

vii) $arg (z)+arg (\bar{z} )$ $(\bar{z}\neq0)$ is ....

viii) If $|z+4|\leq 3$ then the greatest and least values of |z+1| are ..... and .....

ix) if $\frac{z-2}{z+2}=\frac{\pi}{6}$ then the locus of z is .......

x) If $|z|=4$ and $arg (z) = \frac{5\pi}{6}$ then z=

Answer:

i) $|az_1 - bz_2|^2 + |bz_1 + az_2|^2$
$=|az_1 |^2+|bz_2 |^2-2Re(az_1.b(\bar{z_2} ) +|bz_1 |^2+|az_2 |^2+2Re(az_1.b\bar{z_2} ) ?)$
$=|az_1 |^2+|bz_2 |^2+|bz_1 |^2+|az_2 |^2=(a^2+b^2 )(|z_1 |^2+|z_2 |^2 )$
ii) $\sqrt{-25}*\sqrt{-9}=5i*3i=15i^2=-15$
iii) $\frac{(1-i)^3}{1-i^3}=\frac{(1-i)^3}{(1-i)(1+i+i^2 )}$
$=\frac{(1-i)^3}{(1+i-1) }= \frac{1+i^2-2i}{}i$
$=\frac{1-1-2i}{i}=-\frac{2i}{i}=-2$
iv) $i+i^2+i^3+...+i^{1000}=0 \left [ \sum_{n=1}^{1000}i^n=0 \right ]$
v) $\frac{1}{1+i}=1*\frac{1-i}{(1+i)(1-i)} =\frac{1}{2} (1-i)$
vi) Let $z_1=x_1+iy_1 \: \: and \: \: z_2=x_2+iy_2$
$z_1+z_2=(x_1+x_2 )+i(y_1+y_2 )$
If $z_1+z_2$ is real then $y_1+y_2=0$
$y_1=-y_2$
$z_2=x_2-iy_1$
$z_2=x_1-iy_1 (\: \: when\: \: x_1=x_2 )$
So, $z_2=\bar{z_1}$
vii) $arg (z)+arg (\bar{z} )$
$If\: \: arg (z)= \theta , then \arg (\bar{z})=-\theta$
$\theta+(-\theta)=0$
viii) $|z+4|\leq 3$
$=|z+4-3|\leq|z+4|+|-3|$
$=|z+4-3|\leq3+3$
$=|z+4-3|\leq6$
The gratest value is 6 and the least value of $|z+1|$ is 0
ix) $\frac{z-2}{z+2}=\frac{\pi}{6}$
Let $z=x+iy$
$\left |\frac{x+iy-2}{x+iy+2} \right |=\left |\frac{(x-2)+iy}{(x+2)+iy} \right |= \frac{\pi}{6}$
$6|(x-2)+iy|=\pi|(x+2)+iy|$
$6\sqrt{(x-2)^2+y^2 }=\pi\sqrt{(x+2)^2+y^2 }$
$36[x^2+4-4x+y^2 ]=\pi ^2[x^2+4+4x+y^2 ]$
$36x^2+144-144x+36y^2=\pi^2 x^2+4\pi^2+4\pi^2 x+\pi^2 y^2$
$(36-\pi^2 ) x^2+(36-\pi^2 )-(144+4\pi^2 )x+144-4\pi^2=0$
which represents an equation of a circle
x) $|z|=4$ and $arg (z) = \frac{5\pi}{6}$
Let $z=x+iy$
$|z|= \sqrt{x^2+y^2} =4$
$x^2+y^2=16$
$arg(z)=\tan^{-1}\frac{y}{x}=\frac{5\pi}{6}$
$\frac{y}{x}=\tan\left ( \pi-\frac{\pi}{6} \right )$
$=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}$
$x=-\sqrt3 y$
$(-\sqrt3 y)^2+y^2=16$
$3y^2+y^2=16$
$4y^2=16$
$y^2=4$
$y=\pm2$
$x=-2\sqrt3$
$z=-2\sqrt3+2i$

Question:26

State True of False for the following:

i) The order relation is defined on the set of complex numbers.
ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction
iii) For any complex number z, the minimum value of |z| + |z – 11 is 1
iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (1,0) and (0, 1)
v) If z is a complex number such that z ≠ 0 and Re(z) = 0, then Im (z²) = 0
vi) The inequality |z – 4| < |z – 2| represents the region given by x > 3.
(vii) Let Z₁ and Z₂ be two complex numbers such that |z, + z₂| = |z₁ j + |z₂|, then arg (z₁ – z₂) = 0.
(viii) 2 is not a complex number.

Answer:

(i) Comparison of two purely imaginary complex numbers is not possible. However, the two purely real complex number can be compared. So, it is false.

(ii) Let $z= x+iy$

$z.i = (x+iy)i =xi= -y$ which rotates at angle of 180. So, it is ‘false’.

(iii) Let $z= x+iy$

$|z|+|z-1|= \sqrt{x^2+y^2} +\sqrt{(x-1)^2+y^2}$

The value of $|z|+|z-1|$ is minimum when $x=0,y=0 \, \: \: i.e.,1$

Hence, it is true.

iv) Let $z= x+iy$

Given that $\left | z-1 \right |=\left |z-i \right |$

$|x+iy-1|=|x+iy-i|$

$|(x-1)+iy|=|x-(1-y)i|$

$\sqrt{(x-1)^2+y^2}=\sqrt{x^2+(1-y)}^2$

$(x-1)^2+y^2=x^2+(1-y)^2$

$x^2-2x+1+y^2=x^2+1+y^2-2y$

$-2x+2y=0 x-y=0$ which is a straight line slope=1

Now, equation of line through the point 1,0and 0,1

$y-0=\frac{1-0}{0-1} (x-1)$

$y=-x+1$ whose slope=-1

Multiplication of the slopes of two lines =-1*1=-1

So, they are perpendicular. Hence, it is true.

v)Let $z= x+iy$ $z\neq0 \: \: and \: \: Re(z)=0$

Since, real part is 0 $x=0, z=0+iy=iy$

$1m(z^2 )=y^2 i^2=-y^2$ which is real Hence, it is False.

vi) $(z-4)<|z-2|$

Let $z= x+iy$

$|x+iy-4|<|x+iy-2|$

$|(x-4)+iy|<|(x-2)+iy|$

$\sqrt{(x-4)^2+y^2 }<\sqrt{(x-2)^2+y^2}$

$(x-4)^2+y^2<(x-2)^2+y^2$

$(x-4)^2<(x-2)$

$x^2+16-8x<x^2+4-4x$

$8x+4x<-16+4$

$-4x<-12$

$x>3$ Hence, it is true.

vii) Let $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$

$|z_1+z_2 |=|z_1 |+|z_2 |$

$|x_1+iy_1+x_2+iy_2 |=|(x_1+iy_1 )+(x_2+y_2 i)|$

$= \sqrt{(x_1+x_2 )^2+(y_1+y_2 )^2 }$

$=\sqrt{x_1^2+y_1^2} +\sqrt{x_2^2+y_2^2 }$

Squaring both sides, we get $(x_1^2+x_2^2+2x_1 x_2+y_1^2+y_2^2+2y_1 y_2$ )

$=x_1^2+y_1^2+x_2^2+y_2^2+2\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$

$=2x_1 x_2+2y_1 y_2=\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$

$=x_1 x_2+y_1 y_2=\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$

Again squaring on both sides we get $x_1^2 x_2^2+x_1^2 y_2^2+2x_1 y_1 x_2 y_2=x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 2x_1 y_1 x_2 y_2=x_1^2 y_2^2+x_2^2 y_1^2$

$x_1^2 y_1^2+x_2^2 y_1^2-2x_1 y_1 x_2 y_2=0$

$(x_1 y_2-x_2 y_1 )^2=0$

$x_1 y_2-x_2 y_1=0$

$x_1 y_2=x_2 y_1$

$\frac{x_1}{y_1}=\frac{x_2}{y_2}$

$\frac{y_1}{x_1}=\frac{y_2}{x_2}$

$arg(z_1)=arg(z_2)=0$ Hence, it is false.

(viii) Since, every real number is a complex number. So, 2 is a complex number. Hence, it is false.

Question:27

COLUMN A	COLUMN B
a) The polar form of $i+\sqrt3$ is	i) Perpendicular bisector of segment joining (-2,0) and (2,0)
b)The amplitude of $-1+\sqrt{-3}$ is	ii) On or outside the circle having centre at (0,-4) and radius 3.
c) If $\left \| z+2 \right \|=\left \| z-2 \right \|$ then locus of z is	iii) $2\pi/3$
d)If $\left \| z+2i \right \|=\left \| z-2i \right \|$ then locus of z is	iv)Perpendicular bisector of segment joining (0,-2) and (0,2)
e) Region represented by $\left \| z+4i \right \|\geq 3$ is	v) $2\left ( \cos\frac{\pi}{6}+i\sin\frac{\pi}{6} \right )$
f) Region represented by $\left \| z+4 \right \|\geq 3$ is	vi) On or outside the circle having centre at (-4,0) and radius 3units.
g)Conjugate of $\frac{1+2i}{1-i}$ lies in	vii) First Quadrant
h) Reciprocal of 1-i lies in	viii) Third Quadrant

Answer:

a)Given $z=i+\sqrt3$

Polar form of z $=r[cos\theta+isin\theta]=i+\sqrt3$ .

$r=\sqrt{3+1} =2$

And $\tan \alpha = \frac{1}{\sqrt3}$

$\alpha = \frac{\pi}{6}$

Since $x>0,y>0$

Polar form of z $=2[cos\frac{\pi}{6}+isin\frac{\pi}{6}]$

b) Given that $z=-1+\sqrt3=-1+\sqrt3 i$

Here argument z $=\tan^{-1}\left | \frac{\sqrt3}{-1} \right |= \tan^{-1}\sqrt{3}= \frac{\pi}{3}$

So, $\alpha = \frac{\pi}{3}$

Since, x<0 and y>0 $\theta=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$

c) $|z+2|=|z-2|$

$|x+iy+2|=|x+iy-2|$

$|(x-2)+iy|=|(x-2)+iy|$

$\sqrt{(x+2)^2+y^2} =\sqrt{(x-2)^2+y^2}$

$(x+2)^2+y^2=(x-2)^2+y^2$

$(x+2)^2=(x-2)^2$

$x^2+4+4x=x^2+4-4x$

$8x=0 , x=0$

which represents equation of-y axis and is perpendicular to the line joining the points (-2,0) and (2,0)

d) $|z+2i|=|z-2i|$

$(x+iy+2i)=|x+iy-2i|$

$|x+(y+2)i|=|x+(y-2)i|$

$\sqrt{x^2+(y+2)^2} =\sqrt{x^2+(y-2)^2 }$

$x^2+(y+2)^2=x^2+(y-2)^2$

$(y+2)^2=(y-2)^2$

$y^2+4+4y=y^2+4-4y$

$8y=0 ,y=0$

which is the equation of x-axis and is perpendicular to the line segment joining (0,-2)and (0,2)

e) $|z+4i|\geq3$

$|x+iy+4|\geq3$

$|x+(y+4)i|\geq3$

$\sqrt{x^2+(y+4)^2} \geq3$

$x^2+(y+4)^2\geq9$

$x^2+y^2+8y+16 \geq9$

$x^2+y^2+8y+7 \geq0$ $r=\sqrt{(4)^2-7}=3$

which represents a circle on or outside having centre (0,-4)

f) $|z+4|\leq3$

Let $z=x+iy$ Then, $|x+iy+4|\leq3$

$|(x+4)+iy|\leq3$

$\sqrt{(x+4)^2+y^2 } \leq3$

$x^2+8x+16+y^2\leq9$

$x^2+y^2+8x+7 \leq0$ which is a circle having centre (-4,0) and $r=\sqrt{(4)^2-7}=\sqrt9=3$ and is on the circle.

g) Let $z=\frac{1+2i}{1-i}$

$\frac{1+2i}{1-i}* \frac{1+i}{1+i}=\frac{1+i+2i+2i^2}{1-i^2}$

$=\frac{1+i+2i-2}{1+1}= \frac{-1+3i}{2}$

$=-\frac{1}{2}+\frac{3}{2}i$ which lies in third quadrant..

h) Given that z=1-i

Reciprocal of $z=\frac{1}{z}=\frac{1}{1-i}*\frac{1+i}{1+i}$

$=\frac{1+i}{1+i^2}$ which lies in first quadrant.

$So,(a)-(v),(b)-(iii),(c)-(i),(d)-(iv),(e)-(ii),(f)-(vi),(g)-(viii),(h)-(vii)$

Question:28

What is the conjugate of $\frac{2-i}{\left (1-2i \right )^2}$ ?

Answer:

$\frac{2-i}{\left (1-2i \right )^2}$ $= \frac{2-i}{1+4i^2-4i}$
$= \frac{2-i}{-3-4i}= \frac{2-i}{-3-4i}*\frac{-3+4i}{-3+4i}$
$=\frac{-6+8i+3i-4i^2}{9-16i^2}$
$\frac{-6+11i+4}{9-16i^2}=\frac{-2+11i}{9+16}$
$\frac{-2+11i}{25}=\frac{-2}{25}+\frac{11}{25}i$
Hence, $\bar{z}=-\frac{2}{25}-\frac{11}{25}i$

Question:29

If $|z_1| = |z_2|$ , is it necessary that $z_1 = z_2$ ?

Answer:

Let $z_1=x_1+iy_1 \: \: and\: \: z_2=x_2+iy_2$
$|x_1+iy_1 |=|x_2+iy_2 |$
$= \sqrt{x_1^2+y_1^2} = \sqrt{x_2^2+y_2^2}$
$x_1^2+y_1^2=x_2^2+y_2^2$
$\\$Consider $ z_1 = 3 + 4i \\ z_2 = 4 + 3i$
$z_1\neq z_2$ Hence, it is not necessary that $z_1= z_2$

Question:31

Find z if $|z| = 4$ and $arg (z) = 5\pi/6$ ..

Answer:

Polar form of $z=r[cos \theta+i\sin \theta]$
$=4\left [ \cos \frac{5\pi}{6}+i\sin \frac{5\pi}{6} \right ]$
$=4\left [ \cos\left ( \pi-\frac{\pi}{6} \right )+i\sin \left ( \pi-\frac{\pi}{6} \right ) \right ]$
$=4\left [ -\cos\left (\frac{\pi}{6} \right )+i\sin \left (\frac{\pi}{6} \right ) \right ]$
$=4\left [-\frac{\sqrt3}{2}+i\frac{1}{2} \right ]$
$=-2\sqrt3+2i$

Question:32

Find $(1+i)\frac{(2+i)}{(3+i)}$ .

Answer:

$\left | (1+i)\frac{(2+i)}{(3+i)}*\frac{3-i}{3-i} \right |=\left | (1+i).\frac{6-2i+3i-i^2}{9-i^2} \right |$
$=\left | \frac{(1+i)(7+i)}{9+1} \right |$
$=\left | \frac{7+i+7i+i^2}{9+1} \right |$
$=\left | \frac{7+i+7i+i^2}{10} \right |$
$=\left | \frac{7+8i-1}{10} \right |$
$=\left | \frac{6+8i}{10} \right |$
$=\left |\frac{3}{5}+\frac{4}{5}i \right |$
$=\sqrt{\left (\frac{3}{5} \right )^2+\left (\frac{4}{5} \right )^2}=1$

Question:33

Find principal argument of $(1+i\sqrt3)^2$ .

Answer:

$\begin{array}{c} (1+i \sqrt{3})^{2}=1+i^{2} \cdot 3+2 \sqrt{3} i \\ =1-3+2 \sqrt{3} i=-2+2 \sqrt{3} i \\ \tan \alpha=\left|\frac{2 \sqrt{3}}{-2}\right| \\ \tan \alpha=|-\sqrt{3}|=\sqrt{3} \\ \tan \alpha=\tan \frac{\pi}{3} \\ \alpha=\frac{\pi}{3} \end{array}$

Now Re(z)<0 and image(z)>0 $arg(z)=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$
$\text {Hence, the principal arg}=\frac{2 \pi}{3}$

Question:34

Where does z lie, if $|\frac{z-5i}{z+5i}|=1$ .

Answer:

$\begin{array}{c} \text { Let } z=x+i y \quad\left|\frac{x+y i-5 i}{x+i y+5 i}\right|=1 \\ |x+(y-5) i|=|x+(y+5) i| \\ x^{2}+(y-5)^{2}=x^{2}+(y+5)^{2} \\ (y-5)^{2}=(y+5)^{2} \\ \qquad \begin{array}{r} y^{2}+25-10 y=y^{2}+25+10 y \\ 20 y=0 \quad y=0 \end{array} \end{array}$

Hence, z lies on x-axis i.e., real axis.

Question:35

$\sin x + i \cos 2x$ and $\cos x - i \sin 2x$ are conjugate to each other for:
A. $x = n\pi$

B. $x = \left ( n+\frac{1}{2} \right )\frac{\pi}{2}$
C. $x = 0$
D. No value of x

Answer:

The answer is the option (c).
$\text{let }z=\sin x+i \cos 2 x \quad \bar{z}=\sin x-i \cos 2 x$

$But \: \: we \: \: are\: \: given\: \: that\: \: \bar{z}=\cos x-i \sin 2 x$ $$$

$\sin x-\operatorname{icos} 2 x=\cos x-i \sin 2x$ Comparing the real and imaginary part, we get

$\sin \mathrm{x}=\cos \mathrm{x}$ \: \: and\: \: $\cos 2 \mathrm{x}=\sin 2 \mathrm{x}$

$\tan \mathrm{x}=1 \text { and } \tan 2 \mathrm{x}=1$
$\tan 2 \mathrm{x}= \frac{2\tan x }{1-\tan^2 x}=1$
The above value is not satisfied by tan x = 1. Hence no value of x is possible.

Question:36

The real value of α for which the expression $\frac{\left ( 1-i\sin\alpha \right )}{1+2i\sin\alpha}$ is purely real is:
A. $\left ( n+1 \right )\frac{\pi}{2}$
B. $\left ( 2n+1 \right )\frac{\pi}{2}$
C. $n\pi$
D. None of these

Answer:

The answer is the option (c).
Let $z=\frac{\left ( 1-i\sin\alpha \right )}{1+2i\sin\alpha}$
$\begin{aligned} &\begin{array}{c} =\frac{(1-i\sin \alpha)(1-2 i \sin \alpha)}{(1+2 i \sin \alpha)(1-2 i \sin \alpha)} \\\\ =\frac{1-2 i \sin \alpha-i \sin \alpha+2 i^{2} \sin ^{2} \alpha}{(1)^{2}-(2 i \sin \alpha)^{2}} \\\\ =\frac{1-3 i \sin \alpha-2 \sin ^{2} \alpha}{1-4 i^{2} \sin ^{2} \alpha}=\frac{\left(1-2 \sin ^{2} \alpha\right)-3 i \sin \alpha}{1+4 \sin ^{2} \alpha} \\\\ \frac{-3 i \sin \alpha}{1+4 \sin ^{2} \alpha}=0 \\\\ \sin \alpha=0 \quad \operatorname{} \alpha=n \pi \end{array}\\ &\text { Hence, c is correct. } \end{aligned}$

Question:37

If z = x + iy lies in the third quadrant, the $\frac{\bar{z}}{z}$ also lies in the third quadrant if
A. x > y > 0
B. x < y < 0
C. y < x < 0
D. y > x > 0

Answer:

The answer is the option (b).
If z lies in the third quadrant,So, $x<0 \: \: and\: \: y<0\: \: , z=x+iy$
$\begin{aligned} \frac{\bar{z}}{z} &=\frac{x-i y}{x+i y}=\frac{x-i y}{x+i y} * \frac{x-i y}{x-i y} \\ &=\frac{x^{2}+i^{2} y^{2}-2 x y i}{x^{2}-i^{2} y^{2}} \\ &=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}-\frac{2 x y}{x^{2}+y^{2}} i \end{aligned}$
When z lies in third quadrant then $\frac{\bar{z}}{z}$ will also be in third quadrant.
$\begin{array}{c} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}<0 \text { and } \frac{2 x y}{x^{2}+y^{2}}>0 \\\\ x^{2}-y^{2}<0 \text { and } 2 x y>0 \\\\ x^{2}<y^{2} \text { and } x y>0 \\\\ \text { So, } x<y<0 \end{array}$
Hence, b is correct..

Question:38

The value of $(z + 3) (\bar{z}+3)$ is equivalent to
A. $|z + 3|^2$
B. |z – 3|
C. $z^2 + 3$
D. None of these

Answer:

The answer is the option (a).
$Let\: \: z=x+iy \: \: So,\: \: (z+3)(\bar{z} +3)=(x+iy+3)(x-iy+3)$
$=\left [ \left ( x+3 \right ) +iy\right ]\left [ \left ( x+3 \right ) -iy\right ]$
$=(x+3)^2-y^2 i^2=(x+3)^2+y^2$
$=|x+3+iy|^2=|z+3|^2$

Question:39

If $\left ( \frac{1+i}{1-i} \right )^{x}=1$ , then
A. $x = 2n + 1$
B. $x = 4n$
C. $x = 2n$
D. $x = 4n + 1$

Answer:

The answer is the option(b).
$\begin{array}{c} \left(\frac{1+i}{1-i}\right)^{x}=1 \\\\ \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{x}=1 \\\\ \left(\frac{1-1+2 i}{1+1}\right)^{x}=1 \\\\ \left(\frac{2 i}{2}\right)^{x}=1(i)^{x}=1 \end{array}$
$x = 4 n, n \epsilon N$
Hence, the correct option is

A real value of x satisfies the equation $\frac{3-4 i x}{3+4 i x}=\alpha-i \beta$ ( $\alpha, \beta \in R$ ) , if $\alpha ^2 +\beta ^2 =$
A. 1
B. –1
C. 2
D. –2

Answer:

The answer is the option (a).

$\begin{aligned} &\text {Given that} \frac{3-4 i x}{3+4 i x}=\alpha-i \beta\\ &=\frac{3-4 i x}{3+4 i x} * \frac{3-4 i x}{3-4 i x}=\alpha-i \beta\\ &\frac{9-12 i x-12 i x+16 i^{2} x^{2}}{9-16 i^{2} x^{2}}=\alpha-i \beta\\ &=\frac{9-16 x^{2}}{9+16 x^{2}}-\frac{24 x}{9+16 x^{2}}=\alpha-i \beta \ldots \ldots .(i)\\ &\frac{9-16 x^{2}}{9+16 x^{2}}+\frac{24 x}{9+16 x^{2}}=\alpha+i \beta \ldots \ldots \ldots(ii)\\\end{aligned}$
Multiplying eqn (i )and (ii )we get
$\left (\frac{9-16 x^{2}}{9+16 x^{2}} \right )^{2}+\left (\frac{24 x}{9+16 x^{2}} \right )^{2}=\alpha^{2}+ \beta^{2}$
$\frac{81+256 x^{4}-288 x^{2}+576x^2}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}$
$\begin{aligned}\frac{81+256 x^{4}+288 x^{2}}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}\\ \frac{\left(9+16 x^{2}\right)^{2}}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}=1 \end{aligned}$

Question:41

Which of the following is correct for any two complex numbers $z_1\: \: and\: \: \: \: z_2$ ?
A. $|z_1 z_2| = |z_1||z_2|$
B. $arg (z_1z_2) = arg (z_1). Arg (z_2)$
C. $|z_1 + z_2| = |z_1|+ |z_2|$
D. $|z_1 + z_2| \geq |z_1| - |z_2|$

Answer:

The answer is the option (a).
$\begin{array}{c} z_{1}=r_{1}\left(\cos \theta_{1}+i \sin \theta_{2}\right) \\ \left|z_{1}\right|=r_{1} \\ z_{2}=r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\ \left|z_{2}\right|=r_{2} \\\\ z_1z_2=r_{1} r_{2}\left[\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right] \\ =r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right) \cdot r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\ \end{array}$
$\begin{array}{c} =r_{1} r_{2}\left[\left(\cos \theta_{1} \cos \theta_{2}-\sin \theta_{1} \sin \theta_{2}\right)+i\left(\sin \theta_{1} \cos \theta_{2}+\cos \theta_{1} \sin \theta_{2}\right)\right] \\ =r_{1} r_{2}\left[\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right]\left|z_{1} z_{2}\right|=\left|z_{1} \| z_{2}\right| \end{array}$

Hence, option a is correct.

Question:42

The point represented by the complex number $2 -i$ is rotated about origin through an angle $\pi/2$ in the clockwise direction, the new position of point is:
A. $1 + 2i$
B. $-1-2i$
C. $2 + i$
D. $-1 + 2i$

Answer:

The answer is the option (b).
If z rotated through an angle of $\pi/2$ about the origin in clockwise direction .
Then the new position $=z.e^{-\left ( \frac{\pi}{2} \right )}$
$=\left (2-i \right ).e^{-\left ( \frac{\pi}{2} \right )}$
$=\left (2-i \right ).\left [ \cos\left (- \frac{\pi}{2} \right )+isin\left (- \frac{\pi}{2} \right ) \right ]$
$=\left (2-i \right ).\left (0-i \right )$
$=-1-2i$
Hence, the correct option is (b)

Question:43

Let $x, y \in R$ , then $x + iy$ is a non-real complex number if:
A. $x = 0$
B. $y = 0$
C. $x \neq 0$
D. $y \neq 0$

Answer:

The answer is the option (d).
$x+iy$ is a non real complex number if $y\neq0$ ,

Hence, d is correct

Question:44

If $a + ib = c + id$ , then
A. $a^2 + c^2 = 0$
B. $b^2 + c^2 = 0$
C. $b^2 + d^2 = 0$
D. $a^2 + b^2 = c^2 + d^2$

Answer:

The answer is the option (d).
$a+ib=c+id$
$|a+ib|=|c+id|$
$\sqrt{a^2+b^2} =\sqrt{c^2+d^2}$
Squaring both sides, $a^2+b^2=c^2+d^2$
Hence, d is correct

Question:45

The complex number z which satisfies the condition $\left |\frac{i+z}{i-z} \right |=1$ lies on
A. circle $x^2 + y^2 = 1$
B. the x-axis
C. the y-axis
D. the line $x + y = 1$

Answer:

The answer is the option (b).
$\begin{array}{c} \text { Let } z=x+i y \quad\left|\frac{i+x+i y}{i-x-y i}\right|=1 \\\\ \left|\frac{x-(y+1) i}{-x-(y-1) i}\right|=1 \\\\ |x+(y+1) i|=|-x-(y-1) i| \\\\ \sqrt{x^{2}+(y+1)^{2}}=\sqrt{x^{2}+(y-1)^{2}} \\\\ x^{2}+(y+1)^{2}=x^{2}+(y-1)^{2} \\\\ y^{2}+2 y+1=y^{2}-2 y+1 \\ 2 y=-2 y \\\\ y=0 ; x-a x i s \end{array}$

Hence, b is correct

Question:46

If z is a complex number, then
A. $|z^2| > |z|^2$
B. $|z^2| \: \: equals\: \: |z|^2$
C. $|z^2| < |z|^2$
D. $|z^2| \geq |z|^2$

Answer:

. The answer is the option (b).
$Let \: \: z=x+iy \: \: \: \: ,|z|^2=x^2+y^2$
$Now,z^2=x^2+y^2 i^2+2xyi$
$z^2=x^2-y^2+2xyi$
$|z|^2=\sqrt{(x^2-y^2 )^2+(2xy)^2 }$
$=\sqrt{x^4+y^4+2x^2 y^2} =\sqrt{(x^2+y^2 )^2}$
$|z|^2=x^2+y^2=|z|^2$

Hence, b is correct

Question:47

$|z_1 + z_2| = |z_1| + |z_2|$ is possible if
A. $z_1 =\bar{z_1}$

B. z₂₌z₁
C. $arg (z_1) = arg (z_2)$
D. $|z_1| = |z_2|$

Answer:

The answer is the option (c).
Since, $|z_1+z_2 |=|z_1 |+|z_2 |$
$|z_1+z_2 |=|r_1 \left ( \cos \theta _1+i\sin\theta _1 \right )+r_2 \left ( \cos \theta _2+i\sin\theta _2 \right )|$
$=\sqrt{r_{1}^{2} \cos ^{2} \theta_{1}+r_{2}^{2} \cos ^{2} \theta_{2}+2 r_{1} r_{2} \cos \theta_{1} \cos \theta_{2}+r_{1}^{2} \sin ^{2} \theta_{1}+r_{2}^{2} \sin ^{2} \theta_{2}+2 r_{1} r_{2} \sin \theta_{1} \sin \theta_{2}}$
So, $\begin{array}{c} =\sqrt{r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)} \\ \left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right| \\\\ \sqrt{r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)}=r_{1}+r_{2} \end{array}$
Squaring both sides,
$\begin{array}{c} r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \\ 2 r_{1} r_{2}-2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=0 \\ 1-\cos \left(\theta_{1}-\theta_{2}\right)=0 \\ \cos \left(\theta_{1}-\theta_{2}\right)=1 \\ \theta_{1}-\theta_{2}=0 \\ \theta_{1}=\theta_{2} \\ \operatorname{so} \arg \left(z_{1}\right)=\arg \left(z_{2}\right) \end{array}$

Question:48

The real value of θ for which the expression $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is a real number is:
A. $n\pi+\frac{\pi}{4}$

B. $n\pi+(-1)^n\frac{\pi}{4}$
C. $2n\pi+\frac{\pi}{2}$
D. none of these

Answer:

$Let \: \: z=\frac{1+i \cos \theta}{1-2 i \cos \theta} =\frac{1+i \cos \theta}{1-2 i \cos \theta} * \frac{1+2 i \cos \theta}{1+2 i \cos \theta}$ On solving we get
$\frac{1+3 i \cos \theta-2 \cos ^{2} \theta}{1-4 i^{2} \cos ^{2} \theta} \\=\frac{1+3 i \cos \theta-2 \cos ^{2} \theta}{1+4 \cos ^{2} \theta}$
$=\frac{\left(1-2 \cos ^{2} \theta\right)+3 i \cos \theta}{1+4 \cos ^{2} \theta}$
If z is real number then
$\frac{3 \cos \theta}{1+4 \cos ^{2} \theta}=0$
$\\3 \cos \theta=0 \\\cos \theta=0$
$\theta=\frac{(2 n+1) \pi}{2}, n \in N$

Question:49

The value of $arg (x)$ when $x < 0$ is:
A. 0
B. $\frac{\pi}{2}$
C. π
D. none of these

Answer:

The answer is the option (c).
$\\ Let\: \: z= -x+0i \: \: and \: \: x<0 \\ |z|=\sqrt{(-1)^2+(0)^2 }=1,x<0$

Since, the point (-x,0)lies on the negative side of the real axis , principal argument (z)=π

Hence, option c is correct

Question:50

If $f(z)=\frac{7-z}{1-z^2}$ where $z = 1 + 2i$ , then $|f(z)|$ is
A. $\frac{\left | z \right |}{2}$

B. $|z|$
C. $2|z|$
D. none of these

Answer:

The answer is the option (a).
Given that $\\ z=1+2i \\ |z|=\sqrt{(1)^2+(2)^2 }=\sqrt5$
Now $f(z)=\frac{7-z}{1-z^2}$
$\begin{aligned} =\frac{7-(1+2 i)}{1-(1+2 i)^{2}} &=\frac{7-1-2 i}{1-1-4 i^{2}-4 i} \\ =\frac{6-2 i}{4-4 i} &=\frac{3-i}{2-2 i} \end{aligned}$
$\begin{aligned} =\frac{3-i}{2-2 i} * \frac{2+2 i}{2+2 i}=\frac{6+6 i-2 i-2 i^{2}}{4-4 i} \\ =\frac{6+4 i+2}{4+4} &=\frac{8+4 i}{8} \\ =1+\frac{1}{2} i & \end{aligned}$
$f(z)=\sqrt{(1)^{2}+\left(\frac{1}{2}\right)^{2}}=\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2}=\frac{|z|}{2}$

Hence, option a is correct

Important Notes from NCERT Exemplar Class 11 Maths Solution chapter 5

Various laws need to be followed and derived by a mathematician when it comes to multiplying the two complex numbers. All of them are named over the features of the specific formula derivation. Ahead, you will also learn about the Modulus and Conjugate of complex numbers and how to solve sums with their help. There is an Argand and Polar representation in the chapter, relating to graphical presentations. A point needs to be derived at the ‘x’ and ‘y’ axis to form an argand or complex plane.

Students, with the help of NCERT Exemplar Class 11 Maths solutions chapter 5, will not face any issues when trying to solve such problems. It will also help them score better in exams.

All the concepts have been covered in NCERT Exemplar solutions for Class 11 Maths chapter 5. By using NCERT Exemplar Class 11 Maths chapter 5 solutions PDF Download function, students can access quality study material that is effectively constructed by experts for the best learning experience.

Main Subtopics in NCERT Exemplar Class 11 Maths Solution Chapter 5

Introduction
Complex Numbers
Algebra of Complex Numbers
Addition of two complex numbers
Difference of two complex numbers
Multiplication of two complex numbers
Division of two complex number
Power of i
The square roots of a negative real number
Identities
The Modulus and the Conjugate of a Complex Number
Argand Plane and Polar Representation
polar representation of a complex number
Quadratic Equations

What will the students learn from NCERT Exemplar Class 11 Maths Solution Chapter 5?

Belonging to the algebraic part of the subject, it is always fun for the students to solve these problems, once they are clear with the concept. NCERT Exemplar Class 11 Maths chapter 5 solutions will help in understanding the said concept and give examples in how to apply them to questions of any kind.

Other than that, in geometric applications, complex numbers are also very useful, for students in their near future. Students, through NCERT Exemplar Class 11 Maths solutions chapter 5, will also learn graphs in an advanced manner with newer placements and some difficult problems, which would make them sharper and clever.

NCERT Solutions for Class 11 Mathematics Chapters

Chapter 1 Sets

Chapter 2 Relations and Functions

Chapter 3 Trigonometric Functions

Chapter 4 Principles of Mathematical Induction

Chapter 6 Linear Inequalities

Chapter 7 Permutations and Combinations

Chapter 8 Binomial Theorem

Chapter 9 Sequences and Series

Chapter 10 Straight lines

Chapter 11 Conic Sections

Chapter 12 Introduction to Three Dimensional Geometry

Chapter 13 Limits and Derivatives

Chapter 14 Mathematical Reasoning

Chapter 15 Statistics

Chapter 16 Probability

Important Topics in NCERT Exemplar Class 11 Maths Solutions Chapter 5

· Class 11 Maths NCERT Exemplar solutions chapter 5 has detailed that complex numbers, their operations, square root, power, identities, representation of complex numbers and quadratic equations are important topics which students should pay extra attention to.

· In NCERT Exemplar Class 11 Maths chapter 5 solutions, students will also learn about quadratic equations. A quadratic equation is an equation where rearrangement of a particular equation is done via a standard form, where x is the unknown number, whereas, a, b and c, are known numbers. An equation with four numbers is said to be quadratic equations.

Check Chapter-Wise NCERT Solutions of Book

Chapter-1	Sets
Chapter-2	Relations and Functions
Chapter-3	Trigonometric Functions
Chapter-4	Principle of Mathematical Induction
Chapter-5	Complex Numbers and Quadratic equations
Chapter-6	Linear Inequalities
Chapter-7	Permutation and Combinations
Chapter-8	Binomial Theorem
Chapter-9	Sequences and Series
Chapter-10	Straight Lines
Chapter-11	Conic Section
Chapter-12	Introduction to Three Dimensional Geometry
Chapter-13	Limits and Derivatives
Chapter-14	Mathematical Reasoning
Chapter-15	Statistics
Chapter-16	Probability

NCERT Exemplar Class 11 Solutions

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Frequently Asked Questions (FAQs)

1. Are NCERT questions important?

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2. Are these solutions detailed?

Yes, our experts have solved every question in the most detail-oriented way in NCERT Exemplar Class 11 Maths solutions Chapter 5.

3. Which topics are added in this chapter?

In this chapter, the complex numbers are explained, their properties, operations on complex numbers along with quadratic equations.

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Yes, the solutions can be downloaded using the NCERT Exemplar Class 11 Maths Chapter 5 solutions PDF Download function, as one can get the offline PDF after clicking on the download link.

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NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Maths Solutions Chapter 5: Exercise - 1.3

Important Notes from NCERT Exemplar Class 11 Maths Solution chapter 5

Main Subtopics in NCERT Exemplar Class 11 Maths Solution Chapter 5

What will the students learn from NCERT Exemplar Class 11 Maths Solution Chapter 5?

NCERT Solutions for Class 11 Mathematics Chapters

Important Topics in NCERT Exemplar Class 11 Maths Solutions Chapter 5

NCERT Exemplar Class 11 Solutions

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